Edexcel PURE 2024 October — Question 4

Exam BoardEdexcel
ModulePURE
Year2024
SessionOctober
PaperDownload PDF ↗
TopicExponential Equations & Modelling
Typelog(y) vs x: convert and interpret
DifficultyModerate -0.3 This is a straightforward exponential modelling question requiring standard techniques: substituting t=0, converting between logarithmic and exponential forms, and differentiation. Part (b) is essentially guided algebra, and part (c) is routine chain rule application. Slightly easier than average due to the scaffolded structure and standard methods throughout.
Spec1.06c Logarithm definition: log_a(x) as inverse of a^x1.06d Natural logarithm: ln(x) function and properties1.06f Laws of logarithms: addition, subtraction, power rules1.06g Equations with exponentials: solve a^x = b1.06i Exponential growth/decay: in modelling context1.07j Differentiate exponentials: e^(kx) and a^(kx)
  1. The number of bacteria on a surface is being monitored.
The number of bacteria, \(N\), on the surface, \(t\) hours after monitoring began is modelled by the equation $$\log _ { 10 } N = 0.35 t + 2$$ Use the equation of the model to answer parts (a) to (c).
  1. Find the initial number of bacteria on the surface.
  2. Show that the equation of the model can be written in the form $$N = a b ^ { t }$$ where \(a\) and \(b\) are constants to be found. Give the value of \(b\) to 2 decimal places.
  3. Hence find the rate of growth of bacteria on the surface exactly 5 hours after monitoring began.
Question 4:
AnswerMarks
4 (a)l o g N = 2  N = 1 0 0
1 0B1
(1)
AnswerMarks
(b)0 .3 5 t+ 2
l o g N = 0 .3 5 t + 2  N = 1 0
AnswerMarks
1 0M1
0 .3 5 t 2
AnswerMarks
 N = 1 0  1 0dM1
t
AnswerMarks
 N = 1 0 0  2 . 2 4A1
(3)
AnswerMarks
(c)d N
5
Rate of growth at 5 hours is = 1 0 0  l n 2 . 2 4  2 . 2 4
AnswerMarks
d tM1
Allow an answer in the range 4530 to 4550 (bacteria per hour)A1
(2)

Total 6

Method 3: A simultaneous approach that finds values but does not show the result
M1: Sets up two simultaneous equations in a and b
E.g. Uses t = 0 , N = 1 0 0  a = 1 0 0 and t = 1 , N = 1 0 2 .3 5  a b 1 = 2 2 3 . 9
dM1: Solves to find values for a and b.
A0: Has not shown the result so cannot be awarded
Common error: Special Case
0.35t 2
log N =0.35t+2 N =10 +10 so
10
a = 1 0 0 and b=2.24
Or even
0.35t 2 t
log N =0.35t+2 N =10 +10  N =1002.24 so
10
a = 1 0 0 and b = 2 .2 4
Such work resulting in the correct values for a and b scores SC 100
(c) Full marks in part (c) can be scored following the SC
M1: Attempts
d
d
N
t
= K  '' 2 . 2 4
5
'' following through on their 2.24.
This cannot be scored from substituting t = 5 into their a b
t
PMT
dN
5
A1: Calculates =100ln2.242.24 and achieves an answer in the range 4530 to 4550 (bacteria per
dt
hour)
Question 4:
--- 4 (a) ---
4 (a) | l o g N = 2  N = 1 0 0
1 0 | B1
(1)
(b) | 0 .3 5 t+ 2
l o g N = 0 .3 5 t + 2  N = 1 0
1 0 | M1
0 .3 5 t 2
 N = 1 0  1 0 | dM1
t
 N = 1 0 0  2 . 2 4 | A1
(3)
(c) | d N
5
Rate of growth at 5 hours is = 1 0 0  l n 2 . 2 4  2 . 2 4
d t | M1
Allow an answer in the range 4530 to 4550 (bacteria per hour) | A1
(2)
Total 6
Method 3: A simultaneous approach that finds values but does not show the result
M1: Sets up two simultaneous equations in a and b
E.g. Uses t = 0 , N = 1 0 0  a = 1 0 0 and t = 1 , N = 1 0 2 .3 5  a b 1 = 2 2 3 . 9
dM1: Solves to find values for a and b.
A0: Has not shown the result so cannot be awarded
Common error: Special Case
0.35t 2
log N =0.35t+2 N =10 +10 so
10
a = 1 0 0 and b=2.24
Or even
0.35t 2 t
log N =0.35t+2 N =10 +10  N =1002.24 so
10
a = 1 0 0 and b = 2 .2 4
Such work resulting in the correct values for a and b scores SC 100
(c) Full marks in part (c) can be scored following the SC
M1: Attempts
d
d
N
t
= K  '' 2 . 2 4
5
'' following through on their 2.24.
This cannot be scored from substituting t = 5 into their a b
t
PMT
dN
5
A1: Calculates =100ln2.242.24 and achieves an answer in the range 4530 to 4550 (bacteria per
dt
hour)
\begin{enumerate}
  \item The number of bacteria on a surface is being monitored.
\end{enumerate}

The number of bacteria, $N$, on the surface, $t$ hours after monitoring began is modelled by the equation

$$\log _ { 10 } N = 0.35 t + 2$$

Use the equation of the model to answer parts (a) to (c).\\
(a) Find the initial number of bacteria on the surface.\\
(b) Show that the equation of the model can be written in the form

$$N = a b ^ { t }$$

where $a$ and $b$ are constants to be found. Give the value of $b$ to 2 decimal places.\\
(c) Hence find the rate of growth of bacteria on the surface exactly 5 hours after monitoring began.

\hfill \mbox{\textit{Edexcel PURE 2024 Q4}}
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